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Average number of photons

  1. May 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculate the mean number of photons in a cavity at temperature T and the mean energy per photon.

    2. Relevant equations
    In the large volume limit, the log of grand canonical partition function is: ##log(Z_g) = \frac{gV}{h^3}\int log(1-e^{(E-\mu)/kT})dp^3 ##, with g - spin degeneracy, E energy, p - momentum and ##\mu## - chemical potential.

    Also the average number of particles is given by ##N = \lambda \frac{\partial log(Z_g)}{\partial \lambda}|_{T,V}##, with ##\lambda = e^{\mu/kT}##

    3. The attempt at a solution

    I calculated N using the above formula and I obtained: ##N = \frac{gV}{h^3}\int \frac{1}{e^{(E-\mu)/kT}-1}dp^3##. This is for a normal Bose-Einstein gas. Now for a photon I take ##\mu = 0## and ##E=h\nu## and I plug in in this equation. Is this correct? If not, how should I proceed?
     
  2. jcsd
  3. May 6, 2017 #2

    Charles Link

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    Homework Helper

    When you use the Bose function for the mean occupancy off a state is ## \frac{1}{e^{\frac{hc}{\lambda k T}}-1} ## the chemical potential ## \mu ## is set to zero for reasons that I'm not entirely sure. ## \\ ## Anyway, the photon mode counting can proceed in a couple of ways, but I like to use a method that F. Reif uses: ## e^{ik_xx}=e^{ik_x (x+L_x)} ##. This means ## k_x L_x =n_x (2 \pi) ## and similarly for y and z. This means the number of photon modes is ## \Delta N=\Delta^3 n=V \frac{\Delta^3 k}{(2 \pi)^3} ##. Then ## \Delta^3 k=4 \pi k^2 \, dk ## and ## k=\frac{2 \pi}{\lambda} ##. There is also a factor of 2 for photon polarization. You can write the expression entirely in terms of a spectral density ## f(\lambda) d \lambda ##. (We counted photon modes. Don't forget to include the Bose factor for mean occupancy of the state(mode).) ## \\ ## You then integrate over ## d \lambda ## from ## 0 ## to ## +\infty ## to get the mean number of photons in the volume ## V ## at temperature ## T ##. ## \\ ## Editing: Question is, is your integral identical to this, with polarization factor ## g=2 ##, and I believe the answer is yes. You can use ## E_p=\frac{hc}{\lambda} ## and work entirely with wavelength. A google of the subject shows the integral that you have can be evaluated in closed form, but it is a somewhat difficult one. The answer they give is ## I=16 \pi \zeta(3) (\frac{kT}{hc})^3 ##, where ## \zeta ## is the Riemann zeta function. ## \zeta(3)=1.202 ## (approximately).
     
    Last edited: May 6, 2017
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